Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{t^2 + 11t + 30}{t^2 + 5t} \div \dfrac{-4t - 24}{t + 1} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{t^2 + 11t + 30}{t^2 + 5t} \times \dfrac{t + 1}{-4t - 24} $ First factor the quadratic. $r = \dfrac{(t + 6)(t + 5)}{t^2 + 5t} \times \dfrac{t + 1}{-4t - 24} $ Then factor out any other terms. $r = \dfrac{(t + 6)(t + 5)}{t(t + 5)} \times \dfrac{t + 1}{-4(t + 6)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (t + 6)(t + 5) \times (t + 1) } { t(t + 5) \times -4(t + 6) } $ $r = \dfrac{ (t + 6)(t + 5)(t + 1)}{ -4t(t + 5)(t + 6)} $ Notice that $(t + 5)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ \cancel{(t + 6)}(t + 5)(t + 1)}{ -4t(t + 5)\cancel{(t + 6)}} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $r = \dfrac{ \cancel{(t + 6)}\cancel{(t + 5)}(t + 1)}{ -4t\cancel{(t + 5)}\cancel{(t + 6)}} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $r = \dfrac{t + 1}{-4t} $ $r = \dfrac{-(t + 1)}{4t} ; \space t \neq -6 ; \space t \neq -5 $